Chapter 7The Mole: A Measurement of

MatterAt the end of this chapter, you should

be able to:

•Describe how Avogadro’s number is related to a mole of any substance

•Calculate the mass of a mole of any substance

The Mole (aka Avagadro’s Number):

6.02 x 1023

The Mole and Avogadro’s Number

SI unit that measures the amount of substance

1 mole = 6.02 x 1023 representative particles

Representative particles are usually atoms, molecules, or formula units (ions)

But Why the Mole?

Just as 12 = 1 dozen, or 63,360 inches = 1 mile,

the mole allows us to count microscopic items

(atoms, ion, molecules) on a macroscopic scale.

So, 1 mole of any substance is a set number of

Items, namely: 6.02 x 1023.

Chemistry = awesome

Examples:Substance Representative

ParticleChemical Formula

Representative Particles in 1.00

mol

Atomic nitrogen

Atom N 6.02 x 1023

Water Molecule H2O 6.02 x 1023

Calcium ion

Ion Ca2+ 6.02 x 1023

SolveSubstance Representative

ParticleChemical Formula

Representative Particles in

1.00 mol

Nitrogen gas

Calcium Fluoride

Sucrose

Carbon

Answers

Nitrogen gas-molecule-N2

Calcium fluoride-formula unit-CaF2

Sucrose-molecule-C12H22O11

Carbon-atom-C

All have 6.02 x 1023 representative particles in 1.00 mol

How many atoms are in a mole?

Determined from the chemical formula

List the elements and count the atoms

Solve for CO2

C - 1 carbon atom

O - 2 oxygen atoms

Add: 1 + 2 = 3

Answer: 3 times Avogadro’s number of atoms

Solve: How many atoms are in a mole of

1. Carbon monoxide – CO

2. Glucose – C6H12O6

3. Propane – C3H8

4. Water – H2O

How many moles of magnesium is 1.25 x 1023 atoms of magnesium?

Refer to page 174 in text

Divide the number of atoms or molecules given in the example by 6.02 x 1023

Divide (1.25 x 1023) by (6.02 x 1023)

Express in scientific notation

Answer = 2.08 x 10-1 mol Mg

Objectives

Use the molar mass to convert between mass and moles of a substance

Use the mole to convert among measurements of mass, volume, and number of particles

Molar mass

Mass (in grams) of one mole of a substanceBroad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular massCan be unclear: What is the molar mass of oxygen?

O or O2 ? - element O or molecular compound O2 ?

Molar Mass

Gram atomic mass (gam) – atomic mass of an element taken from the periodic table

Gram molecular mass (gmm) – mass of one mole of a molecular compound

Gram formula mass (gfm) – mass of one mole of an ionic compound

Can use molar mass instead of gam, gmm, or gfm

Calculating the Molar Mass of Compounds (Molecular and Ionic)

1. List the elements

2. Count the atoms

3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)

4. Add the masses of each element

5. Express to tenths place

What is the molar mass (gfm) of ammonium carbonate (NH4)2CO3?

N 2 x 14.0 g = 28.0 g

H 8 x 1.0 g = 8.0 g

C 1 x 12.0 g = 12.0 g

O 3 x 16.0 g = 48.0 g

Add ________

Answer 96.0 g

Practice Problems1. How many grams are in 9.45 mol of dinitrogen trioxide (N2O3) ?

a. Calculate the grams in one mole b. Multiply the grams by the number of moles

2. Find the number of moles in 92.2 g of iron(III) oxide (Fe2O3).

a. Calculate the grams in one mole b. Divide the given grams by the

grams in one mole

Answers

1. 718 g N2O3 (one mole is 76.0g)

2. 0.578 mol Fe2O3 (one mole is 159.6 g)

Volume of a Mole of Gas

Varies with a change in temperature or a change in pressure

At STP, 1 mole of any gas occupies a volume of 22.4 L

Standard temperature is 0°C

Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm)

22.4 L is known as the molar volume

22.4 L of any gas at STP contains 6.02 x 1023 representative particles of that gas

One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ)

Study Figure 7.13 on page 186

Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)

Objectives

Define the terms

Calculate the percent composition of a substance from its chemical formula or experimental data

Derive the empirical formula and the molecular formula of a compound from experimental data

Terms to Know

Percent composition – relative amounts of each element in a compound

Empirical formula – lowest whole- number ratio of the atoms of an element in a compound

An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?

1. Calculate the total mass

2. Divide each given by the total mass

and then multiply by 100%

3. Check your answer: The

percentages should total 100%

Answer

The total mass is 8.20 g + 5.40 g = 13.60 g

Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%

Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%

Check your answer: 60.3% + 39.7% = 100%

Calculate the percent composition of propane (C3H8)

1. List the elements2. Count the atoms3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table)4. Express each element as a percentage of the total molar mass5. Check your answer

Answer

Total molar mass = 44.0 g/mol

36.0 g C = 81.8%

8.0 g H = 18.2%

Calculate the mass of carbon in 82.0 g of propane (C3H8)

1. Calculate the percent composition using the formula (See previous problem)

2. Determine 81.8% of 82.0 g

Move decimal two places to the

left (.818 x 82 g)

3. Answer = 67.1 g

Calculating Empirical Formulas

Microscopic – atoms

Macroscopic – moles of atoms

Lowest whole-number ratio may not be the same as the compound formula

Example: The empirical formula of hydrogen peroxide (H2O2) is HO

Empirical Formulas

The first step is to find the mole-to-mole ratio of the elements in the compoundIf the numbers are both whole numbers, these will be the subscripts of the elements in the formulaIf the whole numbers are identical, substitute the number 1

Example: C2H2 and C8H8 have an empirical formula of CHIf either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts

What is the empirical formula of a compound that is 25.9% nitrogen

and 74.1% oxygen?

1. Assume 100 g of the compound, so that

there are 25.9 g N and 74.1 g O

2. Convert to mole-to-mole ratio:

Divide each by mass of one mole

25.9 g divided by 14.0 g = 1.85 mol N

74.1 g divided by 16.0 g = 4.63 mol O

3. Divide both molar quantities by the

smaller number of moles

4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O

5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number )2 x 1 mol N = 22 x 2.5 mol O = 5Answer: The empirical formula is N2O5

Determine the Empirical Formulas

1. H2O2

2. CO2

3. N2H4

4. C6H12O6

5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?

Answers

Compound Empirical Formula

1. H2O2 HO

2. CO2 CO2

3. N2H4 NH2

4. C6H12O6 CH2O

5. HCN

Calculating Molecular FormulasThe molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula

The molecular formula may or may not be the same as the empirical formula

Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.

1. Using the empirical formula, calculate the empirical formula mass (efm)

(Use the same procedure used to calculate molar mass.)2. Divide the known molar mass by the efm 3. Multiply the formula subscripts by this value to get the molecular formula

Answer

Molar mass (efm) is 30.0 g

60.0 g divided by 30.0 g = 2

Answer: C2H8N2

The G.U.S. MethodAllows students to organize data in an easy

way. More importantly, makes my grading job

easier. Works like so:

G: the given. Write down all data given in the

problem WITH proper units.

U: the unknown. Write down what you are

looking for AND the unit.

S: solve. Show ALL work WITH units.